D Dmoney1578 New member Joined May 2, 2008 Messages 7 May 2, 2008 #1 Do know how to do this problem Write the following experssion as a sum or difference: sin(3x) cos(2x)
Do know how to do this problem Write the following experssion as a sum or difference: sin(3x) cos(2x)
skeeter Elite Member Joined Dec 15, 2005 Messages 3,204 May 2, 2008 #2 check the problem statement again ... I believe the expression is incomplete as posted. never mind ... I had sin(3x)cos(2x) +/- cos(3x)sin(2x) = sin(3x +/- 2x) on my mind.
check the problem statement again ... I believe the expression is incomplete as posted. never mind ... I had sin(3x)cos(2x) +/- cos(3x)sin(2x) = sin(3x +/- 2x) on my mind.
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 May 2, 2008 #3 Hello, Dmoney1578! Write the following experssion as a sum or difference: .\(\displaystyle \sin(3x)\cos(2x)\) Click to expand... \(\displaystyle \text{We need a product-to-sum identity: }\;\sin A\cos B \:=\:\frac{1}{2}\bigg[\sin(A-B) + \sin(A+B)\bigg]\) \(\displaystyle \text{Therefore: }\;\sin(3x)\cos(2x) \;=\;\frac{1}{2}\bigg[\sin(x) + \sin(5x)\bigg]\)
Hello, Dmoney1578! Write the following experssion as a sum or difference: .\(\displaystyle \sin(3x)\cos(2x)\) Click to expand... \(\displaystyle \text{We need a product-to-sum identity: }\;\sin A\cos B \:=\:\frac{1}{2}\bigg[\sin(A-B) + \sin(A+B)\bigg]\) \(\displaystyle \text{Therefore: }\;\sin(3x)\cos(2x) \;=\;\frac{1}{2}\bigg[\sin(x) + \sin(5x)\bigg]\)